Detection Limits
All instrumental methods have a degree of
noise associated with the measurement
that limits the amount of analyte that can
be detected.
1. Detection Limit is the lowest
concentration level that can be
determined to be statistically different
from an analyte blank.
1
When a graphical display of results is
obtained, the detection limit of the
instrument can be defined as the
concentration of analyte resulting in a
signal that is twice as the peak to peak
noise (the distance between the two
dashed lines in the schematic below).
2
Peak-to-peak noise level as a basis for detection limit.
A “detectable” analyte signal would be 12
divisions above a line drawn through the
average of the baseline fluctuations
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2. Detection Limit is the concentration that
gives a signal three times the standard
deviation of the background signal.
To calculate the detection limit:
a. Find the average of the blank signal
b. Find the standard deviation of the blank
c. Find the net analyte signal
DL =
analyte conc. * 3*s
analyte signal
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Example
A blank solution in a colorimetric analysis resulted
in absorbance readings of 0.000, 0.008, 0.006,
and 0.003. A 1 ppm standard solution of the
analyte resulted in a reading of 0.051. Calculate
the detection limit.
The standard deviation of the four data
points of the blank can be calculated to be
+ 0.0032 and the mean of the blank is
0.004
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The net reading of the standard = 0.051 –
0.004 = 0.047
The detection limit is the concentration
which results in three times the standard
deviation (3 x 0.0032 = 0.0096).
Detection limits = 1 ppm x 0.0096/0.047 =
0.2 ppm
The absorbance reading of the least
detectable concentration = 0.0096 + 0.004
= 0.014
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Stoichiometric Calculations
7
Review of Fundamental
Concepts
Formula Weight
It is assumed that you can calculate the
formula or molecular weights of compounds
from respective atomic weights of the
elements forming these compounds. The
formula weight (FW) of a substance is the
sum of the atomic weights of the elements
from which this substance is formed from.
The formula weight of CaSO4.7H2O is
8
Element
Ca
S
11 O
14 H
FWt
9
Atomic weight
40.08
32.06
11x16.00
14x1.00

The Mole
The mole is the major word we will use
throughout the course. The mole is defined as
gram molecular weight which means that:
Mole
1 mol H2
1 mol O2
1mol O
1mol NaCl
1 mol Na2CO3
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Grams
2.00 g
32.00 g
16.00 g
58.5 g
106.00 g
Assuming approximate atomic weights of 1.00,
16.00, 23.00, 35.5, and 12.00 atomic mass
units for hydrogen, oxygen, sodium, chlorine
atom, and carbon, respectively.
The number of moles contained in a specific
mass of a substance can be calculated as:
mol = g substance/FW substance
The unit for the formula weight is g/mol
11
In the same manner, the number of mmol of
a substance contained in a specific weight
of the substance can be calculated as
mmol = mol/1000
Or,
mmol = mg substance/FW substance
12
The number of mmol of Na2WO4 (FW
= 293.8 mg/mmol) present in 500 mg
of Na2WO4 can be calculated as
? mmol of Na2WO4 = 500 mg/293.8
(mg/mmol) = 1.70 mmol
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The number of mg contained in 0.25 mmol
of Fe2O3 (FW = 159.7 mg/mmol) can be
calculated as
? mg Fe2O3 = 0.25 mmol Fe2O3 x 159.7
(mg/mmol) = 39.9 mg
Therefore, either the number of mg of a
substance can be obtained from its mmols
or vice versa.
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Calculations Involving Solutions
Molarity (M)
Molarity of a solution can be defined as the
number of moles of solute dissolved n 1 L of
solution. This means that 1 mol of solute will be
dissolved in some amount of water and the
volume will be adjusted to 1 L. The amount of
water may be less than 1 L as the final volume
of solute and water is exactly 1 L.
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Calculations Involving Molarity
Molarity = mol/L = mmol/mL
This can be further formulated as
Number of moles = Molarity X volume in Liters,
or
mol = M (mol/L) x VL
Number of mmol = Molarity x volume in mL , or
mmol = M (mmol/mL) x VmL
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Find the molarity of a solution resulting from
dissolving 1.26 g of AgNO3 (FW = 169.9 g/mol)
in a total volume of 250 mL solution.
First find mol AgNO3 = 1.26 g AgNO3 / 169.9 g/mol
= 7.42x10-3 mol
Also you should know that 250 mL = 0.25 L
Now one can calculate the molarity directly from
Molarity = mol/L
M = 7.42x10-3 mol/0.25 L = 0.0297 mol/L
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We can find the molarity directly in one step
using dimensional analysis
? mol AgNO3 / L = (1.26 g AgNO3 / 250 mL)
x ( mol AgNO3/169.9 g AgNO3) x (1000
mL/1L) =0.0297 M
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Let us find the number of mg of NaCl per mL of
a 0.25 M NaCl solution
First we should be able to recognize the molarity
as 0.25 mol/L or 0.25 mmol/mL. Of course, the
second term offers what we need directly
? mg NaCl in 1 mL = (0.25 mmol NaCl/mL) x
(58.5 mg NaCl/mmol NaCl) = 14.6 mg NaCl/mL
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